Question

tv is a midsegment of δrsu. if rs = 4z - 72 and tv = z - 4, what is the value of z? triangle image with vertices u, s, r and midsegment tv z = blank submit

Explanation:

Step1: Apply the midsegment theorem

The midsegment theorem states that the length of a midsegment of a triangle is half the length of the parallel side. So, \( RS = 2\times TV \).
Given \( RS = 4z - 72 \) and \( TV = z - 4 \), we substitute these into the equation: \( 4z - 72=2(z - 4) \).

Step2: Expand the right - hand side

Using the distributive property \( a(b + c)=ab+ac \), where \( a = 2 \), \( b = z \), and \( c=-4 \), we get \( 4z - 72=2z-8 \).

Step3: Subtract \( 2z \) from both sides

\( 4z-2z - 72=2z-2z - 8 \), which simplifies to \( 2z-72=-8 \).

Step4: Add 72 to both sides

\( 2z-72 + 72=-8 + 72 \), so \( 2z=64 \).

Step5: Divide both sides by 2

\( z=\frac{64}{2} \).

Answer:

\( z = 32 \)